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Nick S
Lv 5
Nick S 發問於 Science & MathematicsMathematics · 8 年前

Bounded Variation and Interpolation on Infinitely Many Nodes?

I thought of a fun little exercise in measure theory that turned out to be more interesting than originally anticipated, and I wanted to share.

Prove or provide a counterexample to the following 'conjecture':

Suppose (a_n) is a monotone (increasing) sequence of real numbers contained in the interval (0,1), and suppose that Lim_{n -> infinity} a_n = 1.

Also, suppose (b_n) is a bounded sequence of real numbers.

Then there exists a function f in BV([0,1]) such that f'(a_n) = b_n for all n >= 1, if and only if

Lim_{n -> infinity} b_n

exists.

更新:

I'm not doing this for a class. I just thought it was an interesting question. I have a complete proof of the backwards direction (I broke it into a few simple cases and one not-so-simple case).

The more I look at it, the more I doubt the forward direction is even true. If, perhaps, we assume that

Sum_{n=1 to infinity} |a_{n+1} - a_n| < infinity

then maybe a quick counterexample will rear its ugly head.

更新 2:

Do you mean

b_n = f'(a_n) = sin(1-1/(n+1))?

Because this does have a limit, namely sin(1). Otherwise, would you mind elaborating on your answer a bit here?

Admittedly, the Sine Integral function was a candidate I considered briefly, but I wasn't able to get anything particularly useful from this.

更新 3:

I see what you did. It wasn't that I was making the wrong substitution. I was simply misinterpreting what you had written originally. That does indeed seem to be a counterexample to the forward direction. I was originally thinking one might be able to design a counterexample giving b_n = (-1)^n (which a simple modification of your example would give). Thanks!

1 個解答

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  • 匿名
    8 年前
    最愛解答

    What about:

    f(x) = int_{t=0}^{x} sin(1/(t-1)) dt, if x in [0, 1).

    f(1) = 0.

    Then f'(x) = sin(1/(x-1)) for x in (0,1), and:

    int_{x=0}^{x=1} |f'(x)|dx <= 1.

    So f(x) has bounded variation.

    Now define the increasing sequence:

    a_n = 1 - 1/(n+1) for n in {1, 2, 3, …}

    Define b_n by:

    b_n = f'(a_n) = sin(-(n+1))

    Then {b_n} is a bounded sequence with no limit.

    ***

    @Nick: To get f'(a_n), I am substituting "1 - 1/(n+1)" for "x" in f'(x).

    This indeed leads to sin(-(n+1)). You seem to be incorrectly

    substituting. You are accidentally substituting "1-1/(n+1)" for "1/(x-1)."

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